The system of equations given in the image is:

$\frac{1}{4}x - 2y = -\frac{3}{4}$ $\frac{1}{5}x + \frac{1}{2}y = 12$

Step 1: Clear the Fractions

To make the equations easier to work with, let's clear the fractions by multiplying each equation by the least common multiple (LCM) of the denominators.

For the first equation:

• The LCM of 4 is 4.

Multiply the entire first equation by 4:

$4 \left(\frac{1}{4}x\right) - 4(2y) = 4\left(-\frac{3}{4}\right)$ $x - 8y = -3 \quad \text{(Equation 1)}$

For the second equation:

• The LCM of 5 and 2 is 10.

Multiply the entire second equation by 10:

$10\left(\frac{1}{5}x\right) + 10\left(\frac{1}{2}y\right) = 10(12)$ $2x + 5y = 120 \quad \text{(Equation 2)}$

Step 2: Solve the System

Now, solve the system of equations:

1. $x - 8y = -3$
2. $2x + 5y = 120$

Let's solve by substitution or elimination. We will use the elimination method.

Multiply the first equation by 2:

$2(x - 8y) = 2(-3)$ $2x - 16y = -6 \quad \text{(Equation 3)}$

Now, subtract Equation 3 from Equation 2: $(2x + 5y) - (2x - 16y) = 120 - (-6)$ $21y = 126$ $y = \frac{126}{21} = 6$

Substitute $y = 6$ into Equation 1:

$x - 8(6) = -3$ $x - 48 = -3$ $x = -3 + 48 = 45$

Solution

The solution to the system is: $x = 45, \quad y = 6$

Would you like any additional details, or do you have any questions?

Related Questions:

1. What would happen if you solved the system using the substitution method instead?
2. How can you check if the solution $x = 45$ and $y = 6$ is correct?
3. Can you express the second equation in terms of $y$ and then substitute?
4. What does it mean geometrically to solve a system of linear equations?
5. How would you approach this problem if the equations were nonlinear?

Tip:

When dealing with fractions in equations, it's often helpful to clear them out by multiplying through by the least common multiple of the denominators before solving the system.